Injectivity of $[(S^n, \;x_o), \;(S^n, \;x_o)]^\circ \longrightarrow [S^n, \;S^n]$ I am studying for an exam in a little more advanced topology, and I want to understand why the mentioned map (i.e. forgetting the base points & hence allowing non-pointed homotopy) is an isomorphism (I know why it is onto, already).
In the lecture notes, the argument begins showing that for
$
x_0 \in S^n
$,
$$
(S^n, \; x_0) \simeq \;((S^n, \;x_0) \vee (I, \;0), \;[1])
$$
where as usual, $I = [0,1]$ is the unit interval; $[1]$ denotes the equivalency class of $1$ in the wedge $(S^n, \;x_0) \vee (I, \;0) \overset{\text{(by def.)}} = (S^n\bigsqcup I)/\sim$, the quotient relation being generated by $x_0 \sim 0$.
This part I have understood.
However, I don't see how to use this to show that the map is injective. The notes say something about using that $S^n$ is simply connected. But how does this show that the base-point forgetting map (now using the indicated homotopy)
$$
[(S^n \vee I, \;[1]), \;(S^n, x_0)]^\circ \longrightarrow [(S^n \vee I), S^n]
$$
is injective?
The notes say something about using that $S^n$ is simply connected. I understand that any homotopy between morphisms
$$
f,g:\;(S^n \vee I, \;[1]) \longrightarrow (S^n, \;x_0)
$$
in the category of pairs of spaces induces a loop over $x_0$ by restricting that homotopy to $\{[1]\} \times I \cong I$, and that this probably plays a role, since any such loop is homotopic to a constant ($S^n$ simply connected).
If I'm not mistaken, I also understand, that, using the inclusions
$$
\begin{align}
&\iota_0:\; (S^n, \;x_0) \hookrightarrow (S^n \vee I, 
\;[0])  \\
&\iota_1:\; (I, \;0) \hookrightarrow (S^n \vee I, \;[0])   \\
\end{align}
$$
allows us to write $f$ as a map induced by the two pointed maps
$$
\begin{align}
&f_0 = f \iota_0:\; (S^n, \; x_0) \longrightarrow (S^n, \;x_0) \\
&f_1 = f \iota_1:\; (I, \;0) \longrightarrow (S^n, \;x_0) \\
\end{align}
$$
induced by the universal property of the sum in the category of pointed spaces (similarly, $g$ comes from $g_0$, $g_1$ then).
Now, given $f \simeq g$ by not necessarily pointed homotopy, how do I get a homotopy
$$
H:\; (\; (S^n \vee I) \times I, \;\{[1]\} \times I \;) \longrightarrow (S^n, \;x_0)
$$
?
If the given homotopy would split up nicely in accordance with how we split up $f,\; g$ into $f_0, \;g_0$ and $f_1, \;g_1$ respectively, then maybe we could use that (?). Sorry for the vague formulation in this paragraph, but I'm a little confused from thinking too much about this.
I already see criticism coming all ways for this, but please, do not give me 'Hatcher'-style answers to this. By this I mean: Please be more formal/rigorous/explicit (for instance, I like the book by/style of Tammo tam Dieck, there are also lecture notes by Lazarev on Algebraic Topology, and I believe I once read some notes by Emily Riehl. I also liked what I read from Spanier - all of these authors I know/remember to be 'more formal/rigorous/explicit').
 A: Suppose $f,g:S^n\to S^n$ are homotopic maps. We want to show that they are based homotopic with respect to some base point $x_0\in S^n$. Of course, this doesn't make sense unless we assume that $f$ and $g$ are themselves based. But we are free to assume this, since every map is homotopic to a based map (this is the surjectiveness statement which you've already proved).
So, we have a homotopy $H:S^n\times I\to S^n$ from $f$ to $g$, where $f(x_0)=x_0$ and $g(x_0)=x_0$. If we look at what happens to the basepoint under this homotopy, we get a closed loop $H(x_0,-):\{x_0\}\times I\to S^n$. Since $S^n$ is simply connected, this loop is null-homotopic via a (based) homotopy $G:\{x_0\}\times I\times I\to S^n$.
Now we apply the homotopy extension property for the pair $\{x_0\}\times I\subseteq S^n\times I$ (a closed cofibration). The map $$H\cup G:S^n\times I\times \{0\}\cup \{x_0\}\times I\times I\to S^n$$ extends to a homotopy $F:S^n\times I\times I\to S^n$. This is homotopy from our original homotopy $H=F|_{S^n\times I\times\{0\}}$ to a based homotopy $H'=F|_{S^n\times I\times\{1\}}:S^n\times I\to S^n$. 
